In a previous video, I made a pre-regulating power supply. And while it works fine, and can handle loads up to one amp, testing it can be a bit tedious. You would need several different power resistors to test a supply like this, not to mention the calculations you need to run to ensure you pick the correct resistor and get the correct results. Isn’t there an easier way to simulate a load so that we can more easily test power supplies? Well, yes. In this video I will show you how to create an adjustable load which allows you to specify just how much current you draw from a voltage source.

Let’s start by reviewing the slow method of testing power supplies. You would first pick a voltage on the supply and determine the expected output based on your design and ohm’s law. For example, if we set the supply to five volts, we should get 33 miliamps with a 150 ohm resistor. And we do. If you don’t, then something is wrong with your supply. What about five volts with a 10 ohm resistor? We get 500 milliamps. And so on and so forth. This process works perfectly fine, its just quite a hassle dealing with all of these power resistors.

But how can we create a variable resistance? The first thing that comes to mind are potentiometers, and yes they technically work. The only problem is that they often times cannot handle the load we need to test. For example, if we ran the supply at five volts and set the potentiometer to 5 ohms, we would get 1 amp, with a power of 5 watts. This is higher than what many potentiometers can handle. It gets even worse with higher voltages and currents. So you could try to find a power potentiometer, but there are better options avaliable for setting our load.

Another other option is using a transistor of some sort, with which can drop a certain amount of voltage leading to a certain current. We could use a BJT, such as the 2n3904, to drop the voltage for us. It would be setup in a voltage follower configuration, and the bottom resistor would control the current based on that voltage setting. However, the 2n3904 has a maximum current of 200mA, which I plan to exceed. The other, more conventient option is to use a MOSFET. That isn’t to say that you can’t use a BJT, just make sure to pick your parts correctly. I will use the IRLZ44N MOSFET.

To really make use of the MOSFET we can use op-amps. If you have watched my previous video you will remember that op-amps will do anything to keep their inputs equal. Let’s imagine for a second a black box where the current flowing through it will generate a voltage feedback to the op-amp which will maintain that same current as compared to the setting made from an adjustable voltage divider. Now let’s actually make this black box.

Let’s start with the first step, which is generating a voltage based on the current flowing. The easiest way to do that is with a current sense resistor. This resistor will generate a voltage based on its current, you will use ohm’s law here. We should pick a small resistor for this purpose, to minimise the effect that it has on the circuit, so I choose a 100 milliohm resistor. This will cause a voltage drop that is one tenth of the current flowing.

Now that we have the voltage drop, we need a way to measure it. We can use another op-amp for this purpose. Since we need two op-amps in total, I choose the LM358 IC which has two op-amps inside. Also keep in mind that the circuit will use a separate nine volt power supply. To measure the drop we will use the differential amplifier configuration. I go into more detail in my op-amp video, but the basic idea is that the differential amplifier outputs the voltage difference between two points, exactly what we said we need to measure.

However, since the resistor is small, the voltage drop will be small aswell. That’s why we need to amplify the difference to match with our voltage divider’s range. We will be using a voltage divider range of 0 to nine volts, so our differential amplifier’s output should be in that range. The current should be between 0 and 2 amps aswell. Keeping those values in mind we can select a resistor for our feedback to get a specific amplification, using 1k for the other resistors. The gain equation for a differential amplifier is the feedback resistor divided by the input resistor. We need to equate our maximum current, which is 2 amps, to our maximum voltage setting, which is nine volts. So, when 2 amps are flowing, we need to get nine volts out. With 2 amps, we will get a drop of 0.2 volts. We need to multiply 0.2 by 45 to get 9 volts. So, we will need to use a resistor that is 45 times larger than the 1k input resistor, which is 45k.

As a side note, make sure to put on a suitable heatsink so that the mosfet doesn’t burn itself out. You can watch my previous video on heatsinks if you want more information.

Anyways, let give it a quick test on the previous power supply. As we can see, we can dynamically adjust the current draw until the supply shuts off. The only thing to do now is solder it together and make it a permanent tool that we can use.

After selecting a perfboard, I soldered together all of the parts and added an additional voltage and current meter, so that you can see what exactly the load is drawing. I’d say that this is a very useful tool, and I recommend that you make one as well if you plan on making and testing a power supply. There are some potential improvements such as a constant power mode, but I will leave that for a future video.

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